The number of 7 digit numbers the sum of whose digits is even, is

A number of the seven digits will be of the form.ax1x2x3x4x50,ax1x2x3x4x51,ax1x2x3x4x52,ax1x2x3x4x53,…ax1x2x3x4x59where a∈{1,2,3,…9}and x1,x2,x3,x4,x5∈{0,1,2,3,…9}.Since sum of the digits should be even, therefore,if a+x1+x2+x3+x4+x5 is an even number, then the digit at units place must be 0, 2, 4, 6, 8 and if a+x1+x2+x3+x4+x5 is an odd number, then the digit at units place will be 1, 3, 5, 7, 9.∴ Required number = 9 × 10 × 10 × 10 × 10 × 10 × 5                                = 9 × 105 × 5 = 45 × 105.